4x+5y=6 and 6x-7y=-20

The center of mass of the lamina with the given density p=5 and the shape shown in the image is (0, -0.529).

To calculate the moments Mx and My:

We first need to identify the functions f(x) and g(x) that define the boundaries of the lamina. In this case, we have:

[tex]f(x) = \sqrt{(1 - x^2)}[/tex]

g(x) = -2

We then use the following formulas to calculate the moments Mx and My:

[tex]M_x = \frac{1}{2} p \int (f(x)^2 - g(x)^2) dx\\M_y = p \int x \times (f(x)) dx[/tex]

where p is the density of the lamina.

Substituting the values of f(x), g(x), and p into the above formulas, we get:

[tex]M_x = \frac{1}{2} \times 5 \times \int ({\sqrt{(1 - x^2}})^2 - (-2)^2) dx\\\\M_x = \frac{1}{2} \times 5 \times \int (1 - x^2 + 4) dx\\\\M_x = \frac{1}{2} \times 5 \times (x - \frac{x^3}{3} + 4x) dx\\\\M_x = \frac{5}{2} (x^2 - \frac{x^4}{3} + 8x) |_{-2}^1\\\\M_x = \frac{5}{2} ((1 - \frac{1}{3} + 8) - (-4 - \frac{16}{3} - 16))\\\\M_x = -\frac{50}{3}\\\\M_y = 5 \times \int x \times ({\sqrt{(1 - x^2)}}) dx[/tex]

This integral is difficult to solve analytically, so we can use numerical methods to approximate its value. Using the trapezoidal rule, we get:

[tex]M_y \approx 5 \times 6.3 \approx 31.2[/tex]

To calculate the center of mass (x_bar, y_bar):

We use the following formulas to calculate the center of mass (x_bar, y_bar):

[tex]x_{bar} = \frac{M_y}{M}\\\\y_{bar} = \frac{M_x}{M}[/tex]

where M is the total mass of the lamina, which is given by:

M = p * A

where A is the area of the lamina.

In this case, the area of the lamina is given by:

[tex]A = \int_{-2}^1 {\sqrt{(1 - x^2)}} dx[/tex]

This integral is also difficult to solve analytically, so we can use numerical methods to approximate its value. Using the trapezoidal rule, we get:

A ≈ 6.3

Therefore, the total mass of the lamina is:

M = p * A = 5 * 6.3 ≈ 31.5

Substituting the values of [tex]M_x, M_y[/tex], and M into the formulas for x_bar and y_bar, we get:

[tex]x_{bar} = \frac{M_y}{M} = \frac{31.2}{31.5} \approx 0\\\\y_{bar} = \frac{M_x}{M} = -\frac{50}{3} :31.5 \approx - 0.529[/tex]

Therefore, the center of mass of the lamina is (0, -0.529).

Answer:

D. [tex]x\geq6[/tex]

Step-by-step explanation:

We are asked to find the inequality that represents an unknown number x is no less than 6.      

Since we know that no less than 6 means greater than or equal to 6.

So our unknown number x should be greater than or equal to 6.

We can represent this information in an inequality as:

[tex]x\geq6[/tex]

Upon looking at our given choices we can see that option D is the correct choice.

The correct answer for the product of [tex](3x+7y)(3x-7y)[/tex]  is  [tex]9x^2+49y^2[/tex].The correct option is [tex]b.[/tex].

Given expression:

[tex](3x+7y)(3x-7y)[/tex]

To simplify multiply first term and second term of  first expression with second expression separately:

Simplify:

[tex]3x(3x+7y)+7y(3x-7y)[/tex]

[tex]= 9x^2+21xy +21xy-49y^2[/tex]

Add Like terms:

[tex]= 9x^2+42xy-49y^2[/tex]

The product of [tex](3x+7y)(3x-7y)[/tex] is [tex]= 9x^2+42xy-49y^2[/tex] . The correct option is [tex]b.[/tex]

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The number of food items Homeroom B needs to collect to have more food items per student than Homeroom A is 16

Food items collected by homeroom A:

Canned food = 30

Dry food = 42

Additional food items = 21

Number of students = 24

Food items per students = (30 + 42 + 21) / 24

= 93/24

Food items collected by homeroom B:

Canned food = 22

Dry food = 24

Additional food items = x

Number of students = 16

Food items per students = (22 + 24 + x) / 16

= (46 + x) / 16

Number of more food items Homeroom B need to collect to have more items per student than Homeroom A:

(46 + x) / 16 > 93/24

cross product

(46 + x) × 24 > 16 × 93

1,104 + 24x > 1,488

24x > 1,488 - 1,104

24x > 384

Divide both sides by 24

x > 384/24

x > 16

Hence, homeroom B should collect more than 16 food item to have a greater item per student

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Homeroom B needs to collect at least 17 more items to exceed Homeroom A in items per student. The calculation is based on the total food items and the number of students in each homeroom.

To find out how many more items Homeroom B needs to collect to have more items per student than Homeroom A, we will first calculate the items per student for each homeroom, considering that Homeroom A collects 21 additional items.

First, calculate the total items of food collected by each homeroom:

Homeroom A:

- Canned food: 30 items

- Dry food: 42 items

- Additional items: 21 items

Total items for Homeroom A:

30 + 42 + 21 = 93 items

Number of students in Homeroom A: 24

Items per student in Homeroom A:

[tex]\[\frac{93}{24} \approx 3.875 \text{ items per student}\][/tex]

Homeroom B:

- Canned food: 22 items

- Dry food: 24 items

Total items for Homeroom B:

22 + 24 = 46 items

Number of students in Homeroom B: 16

Items per student in Homeroom B:

[tex]\[\frac{46}{16} = 2.875 \text{ items per student}\][/tex]

Determine additional items needed for Homeroom B:

Let ( x ) be the additional items needed for Homeroom B to exceed Homeroom A's items per student.

The new total items for Homeroom B would be:

46 + x

The new items per student for Homeroom B would be:

[tex]\[\frac{46 + x}{16}\][/tex]

We need this to be greater than Homeroom A's items per student:

[tex]\[\frac{46 + x}{16} > 3.875\][/tex]

Solving for ( x ):

46 + x > 3.875 × 16

46 + x > 62

x > 62 - 46

x > 16

Thus, Homeroom B needs to collect more than 16 additional items to have more items per student than Homeroom A.

In conclusion, Homeroom B needs to collect at least 17 more items to exceed Homeroom A in items per student.

Answer:

[tex]E=\frac{9r}{i}[/tex]

Step-by-step explanation:

We have been given an equation [tex]Ei=9r[/tex], where [tex]E[/tex] represents ERA, [tex]i[/tex] represents number of innings pitched, and [tex]r[/tex] represents number of earned runs allowed.

To solve the given equation for [tex]E[/tex], we need to separate [tex]E[/tex] on one side on equation.

To separate [tex]E[/tex] on one side on equation, we will divide both sides of equation by [tex]i[/tex].

[tex]\frac{Ei}{i}=\frac{9r}{i}[/tex]

[tex]E=\frac{9r}{i}[/tex], where [tex]i\neq 0[/tex]

Therefore, our required equation would be [tex]E=\frac{9r}{i}[/tex].

Answer: true.

Step-by-step explanation: Here we must see if the use of a compass can be replaced with the use of a tracing paper.

Now, suppose you have a sheet of paper in your table, you can apply pressure in it with your finger at some point, now you can drill a little hole with the point of a pencil at some distance of your finger. Now, with the pencil in place, you can start to rotate it and you will see that an almost perfect circle will appear.

So yes, you can do this, so it's true.

Answer:

Factors are (x +1) ( x -7) .

Step-by-step explanation:

Given  :  x² – 6x – 7.  

To find : Factor.

Solution : We have given  x² – 6x – 7.  

On factoring

x² – 7x + 1x – 7.  

On taking x common from first two terms and 1 from last two terms.

x ( x -7) +1( x -7)

On grouping

(x +1) ( x -7)

Factors are (x +1) ( x -7) .

Therefore, Factors are (x +1) ( x -7) .

For this case we have the following polynomials:

[tex] -x ^ 2 + 9

-3x ^ 2 - 11x + 4
[/tex]

Adding the polynomials we have:

[tex] (-x ^ 2 + 9) + (-3x ^ 2 - 11x + 4)
[/tex]

Rewriting we have:

[tex] x ^ 2 (-1-3) - 11x + (9 + 4)
[/tex]

Therefore, the result of the sum is:

[tex] -4x ^ 2 - 11x + 13
[/tex]

Answer:

the sum of the polynomials is:

[tex] -4x ^ 2 - 11x + 13
[/tex]

option 2

The magnitude of the size change of the figures formed by these matrices is:

                                         2

We are given a  first matrix as:

           [3   -2   1]

and second matrix is given by:

            [6   -4   2]

Since we could clearly observe that when we multiply first matrix by a constant 2 then we obtain a second matrix.

   i.e.   2[3 -2 1]=[2×3   2×(-2)   2×1]

                        = [6   -4   2]

   Hence, the change in the magnitude of the size of two matrices is:

                             2

An employee does not need to stay a specific number of years to make a salary of over $35,000 per year; instead, they must complete a particular level of education. Once they have completed high school, they can earn a median annual income of $40,612, which exceeds the $35,000 threshold.

To answer the question: What is the minimum number of years an employee would have to stay to make a salary of over $35,000 per year? we will need to consider the wage data provided.

According to the information, with further education, the median weekly earnings significantly increase:

Therefore, an individual needs to complete at least a high school diploma to earn an annual salary above $35,000. No additional years of employment beyond obtaining the appropriate level of education are needed to reach over $35,000 in annual salary considering the provided median incomes for each educational level.

The quadratic equation for the word problem, written in standard form is equal to: D. x² + 5x - 104 = 0.

Given the following data:

L = (W + 5) inches.

Area = 104 square inches.

Note: Let the width be x.

Mathematically, the area of a rectangle is calculated by using this formula:

LW = A

(x + 5)x = 104

x² + 5x = 104

x² + 5x - 104 = 0

In Mathematics, the standard form of a quadratic equation is given by ax² + bx + c = 0  ⇔  x² + 5x - 104 = 0.

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To sketch the graph of the inequality y > -2x - 2, draw a dashed line for y = -2x - 2 and shade the area above this line, as this area represents the solution set of the inequality.

To sketch the graph of the linear inequality y > -2x - 2, you would begin by drawing the line y = -2x - 2 as if it were an equation. This line serves as the boundary between the solutions of the inequality and the non-solutions. Since the inequality is greater than, not greater than or equal to, you'll use a dashed line to indicate that points on the line are not included in the solution set.

Next, since the inequality is y > -2x - 2, you will shade the area above the line to indicate that all the points in this area satisfy the inequality. That shaded region represents all the possible solutions to the inequality. Always remember when graphing inequalities to pick a point, often(0,0) if it's not on the line, and check if it satisfies the inequality to ensure correct shading area.

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